TL;DR

Some add-ons to latest post Brute forcing “The monkey and the coconuts”.

In last post we saw that we could reduce the brute force attack first from 15625 candidates down to 1024, then down to one-fourth i.e. 256. This is already a factor of about 60, but of course we can do more.

In particular, for the basic puzzle we observed that the last division assigns $L$ coconuts to each sailor, where $L$ must be of the form $L = 4k - 1$.

So why stop here then?

Let’s rename $L$ to $L_0$, and let’s move on to the next step in the ladder, i.e. the amount of coconuts that are taken by the last sailor during the preliminar divisions that happen through the night. We will call this value $L_1$ 🙄

On the one hand, in the basic puzzle we can easily conclude that $L_1$ must have the same shape as $L_0$, i.e. be of the form:

\[L_1 = 4 k_1 - 1\]

On the other hand, we also know the exact relation between $L_0$ and $L_1$:

\[L_1 = (5 L_0 + 1)/4 \\ 4L_1 = 5 L_0 + 1 \\ 5 L_0 = 4 L_1 - 1\]

Putting these two together we can express $L_0$ in terms of $k_1$, let’s see where we get:

\[5 L_0 = 4(4 k_1 - 1) - 1 = 16 k_1 - 5\]

Now we can observe that the left hand side is divisible by 5, so the right hand side must be divisible by 5 too. As 16 is not divisible by 5, then $k_1$ MUST be divisible by 5 itself, i.e.:

\[5 L_0 = 16 \cdot 5 k - 5 \\ L_0 = 16 k - 1\]

This last relation is extremely interesting, because it tells us that we can iterate over one-sixteenth of the possible candidates between 1 and 1024, i.e. ranging $1 \leq k \leq 64$. Even better times for a brute force attack from a human!

Well, the trend for the basic puzzle is set anyway… why even stop here? Doing the same for the following steps in the ladder brings us to further restrict the range of candidates:

\[L_2 \rightarrow L_0 = 64 k - 1 \quad 1 \leq k \leq 16 \\ L_3 \rightarrow L_0 = 256 k - 1 \quad 1 \leq k \leq 4 \\ L_4 \rightarrow L_0 = 1024 k - 1 \quad 1 \leq k \leq 1 \\\]

We don’t have to take any further step of course, because there’s no constraint for what comes out of $L_5$.

Wait a minute… the last expression has one single candidate for $k$… giving $L_0 = 1023$ that is precisely the solution to the puzzle!

Well… time and again some reasoning beats brute force!

After better reading the page on The monkey and the coconuts, it turns out that of course this approach was already described as a numerical approach. You can take a look at the page to see how a mix of a similar mechanism and some trial-and-error can be applied with the sieve approach, targeting Williams’s puzzle alternative.

Although, in this case… some brute force required.